On 7/30/2015 10:52 PM, David Ryeburn wrote:
In article , rickman
wrote:

I can't say I agree with your "choking" impedance idea. The coax
connects to the balun in the same way it connects to the antenna.

No, it's really rather different.

You aren't explaining anything.


The balun can have no effect on the impedance of the coax shield.
Just as you ask, "What would *stop* current from flowing on the
braid?" when connected to the antenna what will stop the current
from flowing on the braid when connected to the balun?

Because, properly done, there's nowhere for that current to come from.

Uh, the current comes from the shield inside. The connection from the
shield inside and the shield outside is always there. The current
continues to flow on the inside because of the interaction with the
inner conductor current. On reaching the balun this interaction is
disrupted and current can flow back down the outside of the shield if a
better path is not provided. The balun provides that path by the equal
current requirement. In essence the impedance to the shield current is
virtually zero.


The only thing that will stop the current from flowing on the outside of
the shield when connected to the balun is if the balun presents a much
lower impedance path for the current than does the shield.

No, Kirchhoff's current law comes to the rescue. In fact you want the
(current) balun to have a very high impedance ot common mode current.

I think that is the same as saying the impedance to a differential
current is low which is what I am saying, no?


The only way your suggestion makes sense is if the current actually
comes *from* the antenna and the balun prevents that current from
returning to the feed line.

Now you're close. In one way of looking at it, if you connect a coaxial
feedline directly to a dipole antenna, that's EXACTLY where the current
on the outside of the coax braid comes from -- one side of the antenna!

Why does it come from the antenna? If you disconnect the side of the
antenna connected to the shield, the current will flow back down the
outside of the shield, no? The current comes from the shield inside
path, not the antenna.


I'll try to explain the whole thing in a different way.

You aren't saying anything new. This has all been said before and does
not contradict what I am saying...


First, the current flowing "up" the inner conductor of the coax equals
the current flowing "down" the inside surface of the outer conductor of
the coax, no matter what. This assumes the coax shielding is 100%. The
field between the outside surface of the inner conductor and the inner
surface of the outer conductor is completely contained insisde the coax,
and this requires that those two currents be identical. This much
happens no matter what (if anything) is connected to the top end of the
coax.

Now connect something to the top of the coax, and what happens to those
two (equal) curents? It depends upon what is connected.

Connect a physically small resistor (resistance of the resistor doesn't
matter, and if you want, you can put a physically small inductor or a
physically small capacitor in series with that resistor). Kirchhoff's
current law requires the current into the one end of this load to equal
the current out of the other end of the load. The current into the end
of the load connected to the inner conductor of the coax is the same as
the current flowing up that inner conductor (Kirchhoff's law again). The
same amount of curent flows out of the other end of the load, which is
connected to the shield. So the current flowing down the inside of the
shield equals the current flowing out of the shield's end of the load,
and there is none left to flow down the outside of the braid
(Kirchhoff's law once more). If some current did flow down the outside
of the braid, the curent into one end of the load couldn't equal the
current out of the other end of the laod and Kirchhoff would be unhappy.

Now connect an antenna to the coax, instead of a resistor/reactor load.
There isn't a Kirchhoff's law for antennas. Since the two halves of the
dipole aren't perfectly coupled together, it is quite possible for the
current flowing into the left half of the dipole to be different from
the current flowing out of the right half. Should that happen, the
current flowing out of the right half of the antenna down into the coax
shield (inner plus outer surfaces, some on each) will be different from
the current flowing down just the inside of the shield. (Remember the
current down the inside of the shield HAS to equal the current up the
inner conductor.) Where does the difference come from? The difference is
the current flowing down the OUTSIDE of the shield.

Nothing here indicates any current is coming *from* the antenna which
you say further above. "that's EXACTLY where the current on the outside
of the coax braid comes from -- one side of the antenna!" In fact, you
seem to be saying exactly what I am saying, the current comes *from* the
inside of the shield and some returns on the outside of the shield.

You seem to be talking about current flow *from* the inner conductor and
*into* the outer conductor, so maybe this is not significant terminology
since it is just to indicate the polarity of the current rather than the
source.


Put a choke consisting of some bifilar turns of parallel conductor
feedline, or better yet consisting of a piece of small diameter coax,
wound around a toroid, between the top of the coax feedline and the
antenna feedpoint, and the current flowing up one conductor of the choke
(one wire, or the inner conductor if you use coax in the choke) has to
equal the current flowing down the other conductor of the choke (the
other wire, or the inner surface of the shield of the piece of coax in
the choke). That means the the currents out of one conductor at the top
of the choke and into the other conductor at the top of the choke have
to be equal. If they were different, the difference (common mode
current) would have to flow through the common mode impedance of the
choke. If the choke is a good choke, that common mode impedance will be
high. So such common mode current has to be low. Only to the extent that
it isn't zero, can the current out of the one conductor differ from the
current into the other conductor (at the top end of the choke). So the
choke does its best to "force" the currents at its top end to be equal.
That's why it's called a current balun (not a voltage balun). Even if
the impedances of the two halves of the antenna are different, the
current balun will do its best to make the current into one side of the
antenna equal the current out of the other side.

Since the currents at the BOTTOM of the choke in the two conductors are
equal, the stupid coaxial feedline thinks it is feeding a resistor (or
maybe a resistor and a reactor in series), not an antenna, and behaves
accordingly; there is no current flowing down the outside of the coax
shield because there is nowhere for such a current to come from.

If you give this just a little bit of thought, you will realize that the
impedance to the current flowing from the shield inner side into the
choke has to be much lower than the impedance the shield outer side
presents. That path is still there and if the balun did not present a
significantly lower impedance path, current would still flow on the
shield outside.

You can explain it with balanced currents which is not invalid. But
nothing you have said contradicts what I have said.

--

Rick

On 7/31/2015 1:15 AM, rickman wrote:
On 7/30/2015 10:52 PM, David Ryeburn wrote:
In article , rickman
wrote:

I can't say I agree with your "choking" impedance idea. The coax
connects to the balun in the same way it connects to the antenna.

No, it's really rather different.

You aren't explaining anything.

Now you are being argumentative, Rick. Do you want to continue the
dialog or not?

On 7/31/2015 12:19 PM, John S wrote:
On 7/31/2015 1:15 AM, rickman wrote:
On 7/30/2015 10:52 PM, David Ryeburn wrote:
In article , rickman
wrote:

I can't say I agree with your "choking" impedance idea. The coax
connects to the balun in the same way it connects to the antenna.

No, it's really rather different.

You aren't explaining anything.

Now you are being argumentative, Rick. Do you want to continue the
dialog or not?

That is BS. Saying something is "different" with no explanation is not
"discussion" and is not useful to a "dialog". I'm just pointing that out.

In fact, much of what is being said here is talking past the points I
have made. Rather than try to understand what is going on, most here
seem to just repeat the standard explanation without thinking it
through. One of the references discusses the case of a balun made by
wrapping the coax around a core, but when discussing a separate balun
made of wires they say, "When constructed of twisted-pair line, the
effect on imbalance current is the same and for the same reasons, but
operation is more difficuit to visualize". That is a cop-out. A pair
of wires is very clearly different from a coax.

Anytime you don't wish to reply to my posts you are free to refrain.

--

Rick

In article , rickman
wrote:

On 7/31/2015 12:19 PM, John S wrote:
On 7/31/2015 1:15 AM, rickman wrote:
On 7/30/2015 10:52 PM, David Ryeburn wrote:
In article , rickman
wrote:

I can't say I agree with your "choking" impedance idea. The coax
connects to the balun in the same way it connects to the antenna.

No, it's really rather different.

You aren't explaining anything.

Now you are being argumentative, Rick. Do you want to continue the
dialog or not?

Worse than argumentative. I DID explain, but I can't read for him.

I give up. Between us, his word is last. Anyone else who wishes to show
me where my argument involving resistors, Kirchhoff's law, fields inside
coaxial cables, etc. is incorrect, I'll be glad to read. I'm always
ready to learn. In the meantime I'll connect my current baluns the way
I described, and they'll work the way I described. He can do what he
wants.

David, VE7EZM (now), AF7BZ (now), and W8EZE (1949-1967)

--
David Ryeburn

To send e-mail, change "netz" to "net"

On 7/31/2015 1:55 PM, David Ryeburn wrote:
In article , rickman
wrote:

On 7/31/2015 12:19 PM, John S wrote:
On 7/31/2015 1:15 AM, rickman wrote:
On 7/30/2015 10:52 PM, David Ryeburn wrote:
In article , rickman
wrote:

I can't say I agree with your "choking" impedance idea. The coax
connects to the balun in the same way it connects to the antenna.

No, it's really rather different.

You aren't explaining anything.

Now you are being argumentative, Rick. Do you want to continue the
dialog or not?

Worse than argumentative. I DID explain, but I can't read for him.

I give up. Between us, his word is last. Anyone else who wishes to show
me where my argument involving resistors, Kirchhoff's law, fields inside
coaxial cables, etc. is incorrect, I'll be glad to read. I'm always
ready to learn. In the meantime I'll connect my current baluns the way
I described, and they'll work the way I described. He can do what he
wants.

I replied to your post in detail. The comment above was about the
single line response which was not useful. If you want to take an
attitude fine. But please don't act like I was not communicating. I
explained to you my points and now you choose to complain.

I never said baluns don't work. So your comment about your equipment is
not relevant to the discussion.

--

Rick

Jeff wrote:

I give up. Between us, his word is last. Anyone else who wishes to show
me where my argument involving resistors, Kirchhoff's law, fields inside
coaxial cables, etc. is incorrect, I'll be glad to read. I'm always
ready to learn. In the meantime I'll connect my current baluns the way
I described, and they'll work the way I described. He can do what he
wants.

Conservation of energy also shows that with a perfectly matched dipole
all of the energy applied to it will be radiated, or lost as heat in the
elements. So by extension there can be no power flowing on the coax outer.

Jeff
So, clearly, the energy supplied from the feeder and conducted down the
coax outer nerver reaches the antenna. Just as two resistors in
parallel share the energy supplied by a voltage applies across them.


--
Roger Hayter

Jeff wrote:

On 01/08/2015 10:54, Roger Hayter wrote:
Jeff wrote:

I give up. Between us, his word is last. Anyone else who wishes to show
me where my argument involving resistors, Kirchhoff's law, fields inside
coaxial cables, etc. is incorrect, I'll be glad to read. I'm always
ready to learn. In the meantime I'll connect my current baluns the way
I described, and they'll work the way I described. He can do what he
wants.

Conservation of energy also shows that with a perfectly matched dipole
all of the energy applied to it will be radiated, or lost as heat in the
elements. So by extension there can be no power flowing on the coax outer.

Jeff
So, clearly, the energy supplied from the feeder and conducted down the
coax outer nerver reaches the antenna. Just as two resistors in
parallel share the energy supplied by a voltage applies across them.



No, there will be no energy "supplied from the feeder and conducted down
the coax outer" as it will all be conducted into the antenna and
radiated if the antenna is perfectly balanced. The power flowing up the
coax, when perfectly matched, will only 'see' the antenna load, not the
coax outer as and element in parallel.

Jeff

If it was perfectly balanced before you connected the feeder, it will no
longer be perfectly balanced once the shunt impedance to earth of the
coax braid is connected to *one* side of it. Dare I say that that is
why you need a balun? This asymmetrical shunt impedance is the source
of the whole question.

--
Roger Hayter

On 8/1/2015 9:09 AM, Roger Hayter wrote:
Jeff wrote:

On 01/08/2015 10:54, Roger Hayter wrote:
Jeff wrote:

Conservation of energy also shows that with a perfectly matched dipole
all of the energy applied to it will be radiated, or lost as heat in the
elements. So by extension there can be no power flowing on the coax outer.

Jeff
So, clearly, the energy supplied from the feeder and conducted down the
coax outer nerver reaches the antenna. Just as two resistors in
parallel share the energy supplied by a voltage applies across them.



No, there will be no energy "supplied from the feeder and conducted down
the coax outer" as it will all be conducted into the antenna and
radiated if the antenna is perfectly balanced. The power flowing up the
coax, when perfectly matched, will only 'see' the antenna load, not the
coax outer as and element in parallel.

Jeff

If it was perfectly balanced before you connected the feeder, it will no
longer be perfectly balanced once the shunt impedance to earth of the
coax braid is connected to *one* side of it. Dare I say that that is
why you need a balun? This asymmetrical shunt impedance is the source
of the whole question.

Not trying to criticize anyone here. I just want to make the
observation that much of the theory that is applied to the systems
discussed here are generalization which only apply under certain
conditions. Often these conditions are "ideal" and even in the best of
circumstances won't match reality perfectly... but are good enough.
Then a discussion starts where something outside the assumptions is
explored and the "ideal" assumptions are forgotten. The discussion goes
in many directions because the general rules are applied when instead,
the topic needs to be considered at a more fundamental level.

So look at the assumptions (or conditions) that exist in this case and
see which ones apply and which don't. I'm pretty sure that the
condition of an impedance match at the feed point does not exist among
others.

That was my point when I suggested that at the junction of the coaxial
feed line and the balun the same condition exists of there being a path
for current on the outside of the shield. This can divert current flow
from the inside of the shield unless... the current path through the
balun from the shield is a very low impedance.

In order to just eliminate any idea of the balun "preventing" an
unbalanced current flow back from the antenna, replace the antenna with
a matched resistor. Then there will be no reflection of any kind at the
load and the only point that needs to be considered is the junction of
the feed line and the balun.

--

Rick

Jeff wrote:

No, there will be no energy "supplied from the feeder and conducted down
the coax outer" as it will all be conducted into the antenna and
radiated if the antenna is perfectly balanced. The power flowing up the
coax, when perfectly matched, will only 'see' the antenna load, not the
coax outer as and element in parallel.

Jeff

If it was perfectly balanced before you connected the feeder, it will no
longer be perfectly balanced once the shunt impedance to earth of the
coax braid is connected to *one* side of it. Dare I say that that is
why you need a balun? This asymmetrical shunt impedance is the source
of the whole question.


So if what you are saying is correct then terminating a piece of coax
with a 50 ohm resistor will not cause a perfect match and the load will
not be 50 ohms due to the shunt impedance to earth of the coax braid.

I think not.

Jeff

I agree I was probably wrong, and you are probably correct.

--
Roger Hayter

On 8/2/2015 4:20 AM, Jeff wrote:

No, there will be no energy "supplied from the feeder and conducted down
the coax outer" as it will all be conducted into the antenna and
radiated if the antenna is perfectly balanced. The power flowing up the
coax, when perfectly matched, will only 'see' the antenna load, not the
coax outer as and element in parallel.

Jeff

If it was perfectly balanced before you connected the feeder, it will no
longer be perfectly balanced once the shunt impedance to earth of the
coax braid is connected to *one* side of it. Dare I say that that is
why you need a balun? This asymmetrical shunt impedance is the source
of the whole question.


So if what you are saying is correct then terminating a piece of coax
with a 50 ohm resistor will not cause a perfect match and the load will
not be 50 ohms due to the shunt impedance to earth of the coax braid.

I think not.

Maybe you can draw a diagram and show yourself where the current flows.
The earth connection on the shield outside is in parallel with the
earth connection on the shield inside. You can't have any current flow
on the outside because any current flowing on the shield inside balances
with the current flowing in the center conductor. The resistor requires
the two currents are equal with none left to travel down the shield
outside.

Although I am still waiting for jim to explain his idea that there is no
current flow in the coax. If that is true it will make this a *very*
simple picture.

--

Rick